We have a properly convex domain with a Hilbert metric. We define the distance between two points $$x,y$$ by $$d_{\Omega}(x,y) = log([a:x:y:b])$$. If we have a projective transformation that preseves cross ratios we call it an isometry of $$d_{\Omega}$$.

If we restrict ourselves to matrices with determinant plus or minus one $$SL^{\pm} (\Omega)$$ and then lift to the sphere (since the double cover of a projective space $$RP^{n}$$ is a sphere $$S^n$$) and choose one of the preimages $$\tilde{\Omega}$$ with the property that $$\tilde{\Omega}$$ maps to itself it becomes much simpler to talk about eigen values and classify isometries.

$$A$$ is called elliptic if there exists $$p \in \Omega$$ such that $$Ap=p$$. Here $$p=[v]$$ the equivalence class of some vector i.e. $$[Av]=[v]$$. This implies that $$Av = \lambda v$$ for some $$\lambda > 0$$. Hence fixed points coresspond to eigenvectors of the matrix with positve eigenvalues. $$A$$ is parabolic if every eigenvalue of $$A$$ has modulus $$1$$, but $$A$$ is not elliptic. Finally we say that $$A$$ is parabolic otherwise.

We define the translation length $$t(A)=\inf_{x \in \Omega} d_{\Omega}(x,Ax) \geq 0$$ and also define the $$minset(A)=\{x \in \Omega \mid d_{\Omega}(x,Ax) = t(A)\}$$ of an isometry. In Hyperbolic geometery $$H^{n}$$ $$t(A) > 0$$ if and only if $$A$$ is hyperbolic. In this case points along the geodesic are moved the minimum amount i.e. the minset lies in some geodesic, called the axis. If $$t(A)=0$$ then $$A$$ is elliptic if and only if the minset is nonempty and parabolic (along with a unique point $$p \in \partial \tilde{\Omega}$$ called the parabolic fixed point) if and only if the minset is empty.

Proposition: Let $$A \in SL(\Omega)$$, then $$t(A)=log\vert\lambda / \mu\vert$$ where $$\lambda$$ is an eigenvalue of largest modulus and $$\mu$$ is an eigenvalue of smallest modulus.

Proposition: $$A \in SL(\Omega) \subset SL^{\pm}(n+1,R)$$ is elliptic if and only if $$A$$ is conjugate onto $$O(n+1)$$

Proof: By the definition of elliptic there is some point in the domain $$p \in A$$ that is fixed by the elliptic. Here we have a diffeomorphism so $$dA$$ acts on $$T_p \simeq R^{n}$$. $$d_{\Omega}$$ is a norm on $$T_p(\Omega)$$ specifically a Finsler metric. $$dA$$ is an isometry of $$T_p(\Omega)$$, which implies there exists an inner product $$\beta$$ on $$T_p(\Omega)$$ such that $$dA$$ is an isometry of $$\beta$$. Implies $$dA \in Isom(\beta)=O(n)$$. (We did not finish proof in lecture, but this implies the conclusion apparently.)

$$\bar{\Omega}$$ is homeomorphic to some $$[0,1]^n$$ so any isometry of $$\Omega$$, $$A \in SL(\Omega)$$ it preserves the closure $$A(\bar{\Omega})=\bar{\Omega}$$. The Brower fixed point theorem implies that there exists $$p \in \bar{\Omega}$$ such that $$Ap=p$$.

Case1: The fixed point is in the interior $$p \in \Omega$$ then $$A$$ is an elliptic.

Case2: The fixed point is on the boundary $$p \in \partial \bar{\Omega}$$. (In addition it fixes a hyperplane containing that point on the boundary.)

Define a supporting hyper plane $$H$$ to $$\Omega$$ at $$p \in \partial \bar{\Omega}$$ is a projective hyperplane $$H=P(V)$$ with $$dim(V)=n$$, such that $$p \in H$$ and $$H \cap \Omega = \emptyset$$.

Exercise: Supporting hyperplanes exist.

Proposition: The set of supporting hyperplanes $$\{P(ker \phi) : [\phi] \in CC(RP^n)^*\}$$ is dual to a nonempty compact convex set in $$(RP^n)^*$$.

Proof: $$\Omega \subset RP^n$$ and $$\Omega^* \subset RP^n$$ a linear map is in the dual domain if and only if $$P(ker \phi)$$ misses the interior of the domain, $$\Omega^* = P(\{\phi \in V^* \mid \phi(\Omega) >0\})$$. $$\Omega$$ must be in $$S^n \subset R^{n+1} = V$$ for this to make sense, so we have to lift to the double cover ($$V$$ a vector space).

Corollary: If $$A \in SL(\Omega)$$ fixes a point in the boundary of the ball $$p \in \partial \Omega$$ then there exists a supporting hyperplane $$H$$ to $$\Omega$$ at $$p$$ such that $$A(H)=H$$

Proof: $$H$$ corresponds to a point in a dual compact convex set $$C$$. $$A$$ sends supporting hyperplanes to supporting hyperplanes, so $$A: C \rightarrow C$$. Then by the Brower Fixed Point Theorem there exists $$[\phi] \in C$$ such that $$A[\phi]=[\phi]$$. Therefore $$H=ker \phi$$ is a supporting hyperplanefixed by $$A$$.

Structure of fixed set of $$A \in SL(\Omega)$$: Let $$p=[x]$$ for $$x \in R^{n+1}$$, $$A$$ fixes $$p$$ if and only if there is a positive number $$\lambda$$ such that $$Ax=\lambda x$$. In other word equivalent to $$x$$ being and eigenvector with a positive eigenvalue. Given an eigenvalue $$\lambda>0$$ for $$A$$ we have the eigenspace $$E_\lambda$$. Define the Fixset $$Fix(\lambda, A, \bar{\Omega})= \bar{\Omega} \cap P(E_{\lambda})$$ this will be compact and convex as both $$\bar{\Omega}$$ and $$P(E_{\lambda})$$ are. The set of all fixed points will be $$Fix(A,\Omega) = \bigcup_{\lambda > 0} Fix(\lambda, A, \Omega)$$.

Example: Suppose $$\Omega = \Delta$$ then

$SL(\Omega) = \{ ()~a,b > 0 \} \rtimes \{permutation~matrices\}$

Consider $$A=\{ \begin{pmatrix} 2&0&0 \\ 0 & 2 & 0 \\ 0&0&1/4 \end{pmatrix}$$ then the fixed points we have are $$Fix(\lambda =2,A,\Delta)$$ and $$Fix(\lambda = 1/4,A, \Delta)$$.

Dynamics of $$[A] \in P_{+}GL(n+1,R)$$:

First of all the largest eigenvalues will ‘win’ (i.e. they will blow up the fastest as a matrix is multiplied by itself repeatedly) hence we need only consider the largest eigenvalues. Consider $$A=\{ \begin{pmatrix} 2&0&0 \\ 0 & 2 & 2 \\ 0&0&2 \end{pmatrix}$$ then $$A^n=2^n \{ \begin{pmatrix} 1&0&0 \\ 0 & 1 & n \\ 0&0&1 \end{pmatrix}$$. From this we can see that the largest Jordan block will blow up fastest therefore we will just talk about the dynamics of a single $$k+1 \times k+1$$ Jordan block. The growth of some block is $$A^{n}=\lambda^n (I + nN + \frac{n(n-1)}{2}N^2+ ... + (\frac{n}{k}N^k))$$ the last term wins as $$N \rightarrow \infty$$. We define the power of a Jordan blonck with eigenvalue $$\lambda \in C$$ is $$(\vert\lambda\vert,k+1)$$ and we use the lexicographical order. We get a partial ordering on Jordan blocks from this.

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